Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{-7n - 42}{n^2 - 6n + 5} \div \dfrac{n + 6}{n - 1} $
Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{-7n - 42}{n^2 - 6n + 5} \times \dfrac{n - 1}{n + 6} $ First factor the quadratic. $r = \dfrac{-7n - 42}{(n - 1)(n - 5)} \times \dfrac{n - 1}{n + 6} $ Then factor out any other terms. $r = \dfrac{-7(n + 6)}{(n - 1)(n - 5)} \times \dfrac{n - 1}{n + 6} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ -7(n + 6) \times (n - 1) } { (n - 1)(n - 5) \times (n + 6) } $ $r = \dfrac{ -7(n + 6)(n - 1)}{ (n - 1)(n - 5)(n + 6)} $ Notice that $(n + 6)$ and $(n - 1)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -7(n + 6)\cancel{(n - 1)}}{ \cancel{(n - 1)}(n - 5)(n + 6)} $ We are dividing by $n - 1$ , so $n - 1 \neq 0$ Therefore, $n \neq 1$ $r = \dfrac{ -7\cancel{(n + 6)}\cancel{(n - 1)}}{ \cancel{(n - 1)}(n - 5)\cancel{(n + 6)}} $ We are dividing by $n + 6$ , so $n + 6 \neq 0$ Therefore, $n \neq -6$ $r = \dfrac{-7}{n - 5} ; \space n \neq 1 ; \space n \neq -6 $